Problem 77 - A Graphical Approach to College Algebra

Question

Solve each nonlinear system of equations analytically for all real solutions. $$\begin{aligned} x^{2}-x y+y^{2} &=5 \\ 2 x^{2}+x y-y^{2} &=10 \end{aligned}$$

Step-by-Step Solution

Answer

The real solutions are \(( \sqrt{5}, 0 ), ( -\sqrt{5}, 0 ), ( \sqrt{5}, \sqrt{5} ), (-\sqrt{5}, -\sqrt{5})\).

Step 1: Understand the System

We need to solve the nonlinear equations: \( x^2 - xy + y^2 = 5 \) and \( 2x^2 + xy - y^2 = 10 \). These are quadratic in both variables \( x \) and \( y \), so we'll aim to find a strategic way to express one equation in terms of the other for substitution.

Step 2: Simplify the First Equation

Start with the first equation, \( x^2 - xy + y^2 = 5 \). It doesn't easily factor or lend itself to direct substitution. Thus, we'll need to consider the relationship between \( x \) and \( y \) by comparing both equations.

Step 3: Simplify the Second Equation

Consider the second equation, \( 2x^2 + xy - y^2 = 10 \). We'll use this equation to solve for one variable in terms of the other. Multiply the first equation by 2 for comparison: \( 2(x^2 - xy + y^2) = 10 \), which gives \( 2x^2 - 2xy + 2y^2 = 10 \).

Step 4: Subtract Equations

Now subtract the twice-first equation from the second equation: \[ (2x^2 + xy - y^2) - (2x^2 - 2xy + 2y^2) = 10 - 10 \]This simplifies to \( 3xy - 3y^2 = 0 \), or \( 3y(x-y) = 0 \).

Step 5: Solve for Common Solutions

The equation \( 3y(x-y) = 0 \) gives two possibilities: \( y = 0 \) or \( x = y \). We'll explore both scenarios.

Step 6: Substitute \( y = 0 \) into Original Equations

First, substituting \( y = 0 \) into the first equation:\[ x^2 = 5 \] This gives \( x = \pm\sqrt{5} \).Check in the second equation:\[ 2x^2 = 10 \]This is satisfied for \( x = \pm\sqrt{5} \).

Step 7: Substitute \( x = y \) into Original Equations

Next, substitute \( x = y \) into the first equation:\[ x^2 - x^2 + x^2 = 5 \] Which simplifies to \( x^2 = 5 \), yielding \( x = \pm\sqrt{5} \). Thus, \( y = \pm\sqrt{5} \) as well.

Step 8: Conclude Solutions

The solutions are \(( \sqrt{5}, 0 ), ( -\sqrt{5}, 0 ), ( \sqrt{5}, \sqrt{5} ), (-\sqrt{5}, -\sqrt{5})\). All these solutions satisfy both original equations and are verified upon substitution.

Key Concepts

Quadratic Equations

Understanding quadratic equations is essential when solving systems like these. A quadratic equation is any equation that can be expressed in the form\[-ax^2 + bx + c = 0,\]where \(a\), \(b\), and \(c\) are constants. In our original exercise, both provided equations have terms of the second degree. Quadratic equations come in various forms:

• Standard form: This is the usual \(ax^2 + bx + c = 0\).
• Factored form: This expresses the equation as \((x - p)(x - q) = 0\).
• Vertex form: Where the equation looks like \(a(x - h)^2 + k = 0\).

These forms help identify different properties of the quadratic equation, like roots and the shape of its corresponding parabola. Mastering these variations aids in effectively tackling quadratic systems of equations as seen here.

Substitution Method

The substitution method is a handy tool for solving systems of equations like this one. The idea is to solve one equation for one of the variables, and then substitute that expression into another equation.In our exercise:

• We used the second equation, \(2x^2 + xy - y^2 = 10\), and simplified it to compare with the first equation.
• Multiplying the first equation by 2, we formed \(2x^2 - 2xy + 2y^2 = 10\).
• Subtracting these equations, we isolated terms to find \(3y(x - y) = 0\).

This process reveals possible solutions because it reduces the complexity, resulting in easily solvable forms. After substitution, each possible scenario is explored further, leading us to find \(y = 0\) or \(x = y\). Understanding and applying the substitution method effectively helps simplify nonlinear systems.

Real Solutions

When solving equations, especially nonlinear systems, we often seek real solutions. A real solution is simply a value that satisfies the equation without involving imaginary numbers.In our original system:

• We found real solutions by considering both \(y = 0\) and \(x = y\). For \(y = 0\), substituting back provided \(x = \pm\sqrt{5}\).
• For \(x = y\), we found solutions \((x, y) = (\pm\sqrt{5}, \pm\sqrt{5})\).

Each candidate solution was substituted back into the original equations to verify if they truly satisfied the systems. Upon verification, all were confirmed as real solutions. This exemplifies the process of checking each derived solution within the context of the original problem to ensure comprehensive understanding and correctness.